Symmetric 6 sided dice, exactly 5 sixes; chance of six > 0.95
Hello, I'm having problems with probability and I hope You can help me! (If this is more suitable for uni level, please move it there)
My problem is : A symmetric dice is rolled 30 times. What is the probability that a 6 turns up exactly 5 times?
Obviously cardinality of is .
Now we focus of cardinality of . In order to have exactly 5 sixes, we need to have 5 sixes in and rest can be from 1 to 5. So we have choices for rest. When we selected numbers, we can "move" block of sixes in any way we want (since it doesn't matter if 5 sixes felt in first 5 throws or last), we can do that in 25 ways*. So and we get probability by dividing those two:
*The number of r-permutations of set with n elements is , we can consider here a block of 5 sixes as one big block. By this logic we can choose block positions in , which would be my proof for above selection.
Second problem: How many dice you need to throw in order to have more then 0.95 chance of six.
I do not have a clear idea how to solve this one, my first thought was calculating chance of complement (i.e. chances of numbers 1 to 5 < 0.5), so basically I'm looking at and I get n has to be over 16. My problem is how to solve this in more strict way.
Thank You for Your time.
Re: Symmetric 6 sided dice, exactly 5 sixes; chance of six > 0.95
This is binomial: .
Originally Posted by magicka
Now for this problem: